Question

Phân tích các đa thức sau thành nhân tử:

a) xy – 3x + 2y – 6

b) x²y + 4xy + 4y – y³

c) x² + y² + xz + yz + 2xy

d) x³ + 3x² – 3x – 1

Answer 1

a) xy – 3x + 2y – 6

= (xy - 3x) + (2y - 6)

= x(y - 3) + 2(y - 3)

= (y - 3)(x + 2)

b) x²y + 4xy + 4y – y³

= y(x² + 4x + 4 - y²)

= y[(x² + 4x + 4) - y²]

= y[(x + 2)² - y²]

= y(x + 2 + y)(x + 2 - y)

c) x² + y² + xz + yz + 2xy

= (x² + 2xy + y²) + (xz + yz)

= (x + y)² + z(x + y)

= (x + y)(x + y + z)

d) x³ + 3x² – 3x – 1

= (x³ - 1) + (3x² - 3x)

= (x - 1)(x² + x + z) + 3x(x - 1)

= (x - 1)(x² + 4x + 1)

Answer 2

a ) 

\(xy-3x+2y-6\)

\(=\left(xy+2y\right)-3x-6\)

\(=y\left(x+2\right)-3\left(x+2\right)\)

\(=\left(y-3\right)\left(x+2\right)\)

b ) 

\(x^2y+4xy+4y-y^3\)

\(=y\left(x^2+4x+4-y^2\right)\)

\(=y\left[\left(x+2\right)^2-y^2\right]\)

\(=y\left(x+2-y\right)\left(x+2+y\right)\)

c ) 

\(x^2+y^2+xz+yz+2xy\)

\(=\left(x+y\right)^2+z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y+z\right)\)

Answer 3

d ) 

\(x^3+3x^2-3x-1\)

\(=3x\left(x-1\right)+\left(x^3-1^3\right)\)

\(=3x\left(x-1\right)+\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x-1\right)\left[3x+\left(x^2+x+1\right)\right]\)

\(=\left(x-1\right)\left(4x+x^2+1\right)\)